Resistors in Series and Parallel

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Resistors in Series and

Parallel

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OpenStaxCollege

Most circuits have more than one component, called a resistor that limits the flow of charge in the circuit A measure of this limit on charge flow is called resistance The simplest combinations of resistors are the series and parallel connections illustrated in [link] The total resistance of a combination of resistors depends on both their individual values and how they are connected

(a) A series connection of resistors (b) A parallel connection of resistors.

Resistors in Series

When are resistors in series? Resistors are in series whenever the flow of charge, called the current, must flow through devices sequentially For example, if current flows

through a person holding a screwdriver and into the Earth, then R1in[link](a) could be

the resistance of the screwdriver’s shaft, R2the resistance of its handle, R3the person’s

body resistance, and R4the resistance of her shoes

[link] shows resistors in series connected to a voltage source It seems reasonable that the total resistance is the sum of the individual resistances, considering that the current has to pass through each resistor in sequence (This fact would be an advantage to a person wishing to avoid an electrical shock, who could reduce the current by wearing high-resistance rubber-soled shoes It could be a disadvantage if one of the resistances

were a faulty high-resistance cord to an appliance that would reduce the operating current.)

Three resistors connected in series to a battery (left) and the equivalent single or series

resistance (right).

To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop, in each resistor in[link]

According to Ohm’s law, the voltage drop, V, across a resistor when a current flows through it is calculated using the equation V = IR, where I equals the current in amps (A) and R is the resistance in ohms ( Ω ) Another way to think of this is that V is the voltage necessary to make a current I flow through a resistance R.

So the voltage drop across R1is V1= IR1, that across R2 is V2 = IR2, and that across R3

is V3 = IR3 The sum of these voltages equals the voltage output of the source; that is,

V = V1+ V2+ V3

This equation is based on the conservation of energy and conservation of charge

Electrical potential energy can be described by the equation PE=qV , where q is the electric charge and V is the voltage Thus the energy supplied by the source is qV, while

that dissipated by the resistors is

qV1+ qV2+ qV3

Connections: Conservation Laws

The derivations of the expressions for series and parallel resistance are based on the laws of conservation of energy and conservation of charge, which state that total charge and total energy are constant in any process These two laws are directly involved in all electrical phenomena and will be invoked repeatedly to explain both specific effects and the general behavior of electricity

These energies must be equal, because there is no other source and no other destination

for energy in the circuit Thus, qV = qV1+ qV2+ qV3 The charge q cancels, yielding

V = V1 + V2+ V3, as stated (Note that the same amount of charge passes through the

battery and each resistor in a given amount of time, since there is no capacitance to store charge, there is no place for charge to leak, and charge is conserved.)

Now substituting the values for the individual voltages gives

V = IR1 + IR2+ IR3= I(R1+ R2+ R3)

Note that for the equivalent single series resistance Rs, we have

V = IRs

This implies that the total or equivalent series resistance Rs of three resistors is

Rs = R1+ R2+ R3

This logic is valid in general for any number of resistors in series; thus, the total

resistance Rsof a series connection is

Rs = R1+ R2+ R3+ ,

as proposed Since all of the current must pass through each resistor, it experiences the resistance of each, and resistances in series simply add up

Calculating Resistance, Current, Voltage Drop, and Power Dissipation: Analysis of a Series Circuit

Suppose the voltage output of the battery in [link] is 12.0 V, and the resistances are

R1 = 1.00 Ω , R2= 6.00 Ω , and R3 = 13.0 Ω (a) What is the total resistance? (b) Find the current (c) Calculate the voltage drop in each resistor, and show these add to equal the voltage output of the source (d) Calculate the power dissipated by each resistor (e) Find the power output of the source, and show that it equals the total power dissipated

by the resistors

Strategy and Solution for (a)

The total resistance is simply the sum of the individual resistances, as given by this equation:

Rs =

=

=

R1+ R2+ R3

1.00 Ω + 6.00 Ω + 13.0 Ω

20.0 Ω

Strategy and Solution for (b)

The current is found using Ohm’s law, V = IR Entering the value of the applied voltage

and the total resistance yields the current for the circuit:

I = R Vs = 12.0 V20.0 Ω = 0.600 A

Strategy and Solution for (c)

The voltage—or IR drop—in a resistor is given by Ohm’s law Entering the current and

the value of the first resistance yields

V1 = IR1= (0.600 A)(1.0 Ω ) = 0.600 V

Similarly,

V2 = IR2= (0.600 A)(6.0 Ω ) = 3.60 V

and

V3 = IR3= (0.600 A)(13.0 Ω ) = 7.80 V

Discussion for (c)

The three IR drops add to 12.0 V, as predicted:

V1 + V2+ V3 = (0.600+3.60+7.80) V=12 0 V

Strategy and Solution for (d)

The easiest way to calculate power in watts (W) dissipated by a resistor in a DC circuit

is to use Joule’s law, P = IV, where P is electric power In this case, each resistor has the same full current flowing through it By substituting Ohm’s law V = IR into Joule’s

law, we get the power dissipated by the first resistor as

P1 = I2R1= (0.600 A)2(1.00 Ω ) = 0.360 W

Similarly,

P2 = I2R2= (0.600 A)2(6.00 Ω ) = 2.16 W

and

P3 = I2R3= (0.600 A)2(13.0 Ω ) = 4.68 W

Discussion for (d)

Power can also be calculated using either P = IV or P = V R2, where V is the voltage drop

across the resistor (not the full voltage of the source) The same values will be obtained

Strategy and Solution for (e)

The easiest way to calculate power output of the source is to use P = IV, where V is the

source voltage This gives

P = (0.600 A)(12.0 V) = 7.20 W.

Discussion for (e)

Note, coincidentally, that the total power dissipated by the resistors is also 7.20 W, the same as the power put out by the source That is,

P1 + P2+ P3 = (0.360+2.16+4.68) W = 7.20 W

Power is energy per unit time (watts), and so conservation of energy requires the power output of the source to be equal to the total power dissipated by the resistors

Major Features of Resistors in Series

1 Series resistances add: Rs = R1+ R2+ R3+

2 The same current flows through each resistor in series

3 Individual resistors in series do not get the total source voltage, but divide it

Resistors in Parallel

[link] shows resistors in parallel, wired to a voltage source Resistors are in parallel when each resistor is connected directly to the voltage source by connecting wires having negligible resistance Each resistor thus has the full voltage of the source applied

to it

Each resistor draws the same current it would if it alone were connected to the voltage source (provided the voltage source is not overloaded) For example, an automobile’s headlights, radio, and so on, are wired in parallel, so that they utilize the full voltage of the source and can operate completely independently The same is true in your house, or any building (See[link](b).)

(a) Three resistors connected in parallel to a battery and the equivalent single or parallel resistance (b) Electrical power setup in a house (credit: Dmitry G, Wikimedia Commons)

To find an expression for the equivalent parallel resistance Rp, let us consider the currents that flow and how they are related to resistance Since each resistor in the circuit

has the full voltage, the currents flowing through the individual resistors are I1= R V1,

I2= R V2, and I3= R V3 Conservation of charge implies that the total current I produced by

the source is the sum of these currents:

I = I1 + I2+ I3

Substituting the expressions for the individual currents gives

I = R V

1 + R V

2 + R V

3 = V( 1

R1 + R1

2 + R1

3) Note that Ohm’s law for the equivalent single resistance gives

I = R Vp = V( 1

Rp)

The terms inside the parentheses in the last two equations must be equal Generalizing

to any number of resistors, the total resistance Rp of a parallel connection is related to the individual resistances by

1

Rp = R11 + R12 + R1.3 +

This relationship results in a total resistance Rp that is less than the smallest of the individual resistances (This is seen in the next example.) When resistors are connected

in parallel, more current flows from the source than would flow for any of them individually, and so the total resistance is lower

Calculating Resistance, Current, Power Dissipation, and Power Output: Analysis of a Parallel Circuit

Let the voltage output of the battery and resistances in the parallel connection in[link]

be the same as the previously considered series connection: V = 12.0 V, R1= 1.00 Ω

, R2= 6.00 Ω , and R3= 13.0 Ω (a) What is the total resistance? (b) Find the total current (c) Calculate the currents in each resistor, and show these add to equal the total current output of the source (d) Calculate the power dissipated by each resistor (e) Find the power output of the source, and show that it equals the total power dissipated by the resistors

Strategy and Solution for (a)

The total resistance for a parallel combination of resistors is found using the equation below Entering known values gives

1

Rp = R11 + R12 + R13 = 1.00 Ω1 + 6.00 Ω1 + 13.0 Ω1

Thus,

1

Rp = 1.00Ω + 0.1667Ω + 0.07692Ω = 1.2436Ω

(Note that in these calculations, each intermediate answer is shown with an extra digit.)

We must invert this to find the total resistance Rp This yields

Rp = 1.24361 Ω = 0.8041 Ω

The total resistance with the correct number of significant digits is Rp= 0.804 Ω

Discussion for (a)

Rpis, as predicted, less than the smallest individual resistance

Strategy and Solution for (b)

The total current can be found from Ohm’s law, substituting Rpfor the total resistance This gives

I = R V

p = 0.8041 Ω12.0 V = 14.92 A

Discussion for (b)

Current I for each device is much larger than for the same devices connected in series

(see the previous example) A circuit with parallel connections has a smaller total resistance than the resistors connected in series

Strategy and Solution for (c)

The individual currents are easily calculated from Ohm’s law, since each resistor gets the full voltage Thus,

I1= R V

1 = 1.00 Ω12.0 V = 12.0 A

Similarly,

I2= R V

2 = 6.00 Ω12.0 V = 2.00 A

and

I3= R V

3 = 12.0 V13.0 Ω = 0.92 A

Discussion for (c)

The total current is the sum of the individual currents:

I1+ I2+ I3 = 14.92 A

This is consistent with conservation of charge

Strategy and Solution for (d)

The power dissipated by each resistor can be found using any of the equations relating

power to current, voltage, and resistance, since all three are known Let us use P = V R2, since each resistor gets full voltage Thus,

P1 = V R21 = (12.0 V)1.00 Ω2 = 144 W

Similarly,

P2 = V R22 = (12.0 V)6.00 Ω2 = 24.0 W

and

P3 = V R2

3 = (12.0 V)13.0 Ω2 = 11.1 W

Discussion for (d)

The power dissipated by each resistor is considerably higher in parallel than when connected in series to the same voltage source

Strategy and Solution for (e)

The total power can also be calculated in several ways Choosing P = IV, and entering

the total current, yields

P = IV = (14.92 A)(12.0 V) = 179 W.

Discussion for (e)

Total power dissipated by the resistors is also 179 W:

P1 + P2+ P3 = 144 W+24 0 W+11 1 W=179 W

This is consistent with the law of conservation of energy

Overall Discussion

Note that both the currents and powers in parallel connections are greater than for the same devices in series

Major Features of Resistors in Parallel

1 Parallel resistance is found from R1

p = R1

1 + R1

2 + R1

3 + , and it is smaller than any individual resistance in the combination

2 Each resistor in parallel has the same full voltage of the source applied to it (Power distribution systems most often use parallel connections to supply the myriad devices served with the same voltage and to allow them to operate independently.)

3 Parallel resistors do not each get the total current; they divide it

Combinations of Series and Parallel

More complex connections of resistors are sometimes just combinations of series and parallel These are commonly encountered, especially when wire resistance is considered In that case, wire resistance is in series with other resistances that are in parallel

Combinations of series and parallel can be reduced to a single equivalent resistance using the technique illustrated in [link] Various parts are identified as either series or parallel, reduced to their equivalents, and further reduced until a single resistance is left The process is more time consuming than difficult

This combination of seven resistors has both series and parallel parts Each is identified and reduced to an equivalent resistance, and these are further reduced until a single equivalent

resistance is reached.

The simplest combination of series and parallel resistance, shown in [link], is also the

most instructive, since it is found in many applications For example, R1 could be the

R2 and R3 could be the starter motor and a passenger compartment light We have previously assumed that wire resistance is negligible, but, when it is not, it has important effects, as the next example indicates

Calculating Resistance, IR Drop, Current, and Power Dissipation: Combining Series

and Parallel Circuits

[link] shows the resistors from the previous two examples wired in a different way—a

combination of series and parallel We can consider R1 to be the resistance of wires

leading to R2and R3 (a) Find the total resistance (b) What is the IR drop in R1? (c) Find

the current I2through R2 (d) What power is dissipated by R2?

These three resistors are connected to a voltage source so that R 2 and R 3 are in parallel with one

another and that combination is in series with R 1 .

Strategy and Solution for (a)

To find the total resistance, we note that R2and R3are in parallel and their combination

Rp is in series with R1 Thus the total (equivalent) resistance of this combination is

Rtot = R1+ Rp

First, we find Rpusing the equation for resistors in parallel and entering known values:

1

Rp = R12 + R13 = 6.00 Ω1 + 13.0 Ω1 = 0.2436Ω

Inverting gives

Rp = 0.24361 Ω = 4.11 Ω

So the total resistance is

Rtot = R1+ Rp = 1.00 Ω + 4.11 Ω = 5.11 Ω

The total resistance of this combination is intermediate between the pure series and pure parallel values (20.0 Ω and 0.804 Ω, respectively) found for the same resistors in the two previous examples

Strategy and Solution for (b)

To find the IR drop in R1, we note that the full current I flows through R1 Thus its IR

drop is

V1 = IR1

We must find I before we can calculate V1 The total current I is found using Ohm’s law

for the circuit That is,

I = R V

tot = 5.11 Ω12.0 V = 2.35 A

Entering this into the expression above, we get

V1 = IR1= (2.35 A)(1.00 Ω ) = 2.35 V

Discussion for (b)

The voltage applied to R2 and R3 is less than the total voltage by an amount V1 When wire resistance is large, it can significantly affect the operation of the devices

represented by R2and R3

Strategy and Solution for (c)

To find the current through R2, we must first find the voltage applied to it We call

this voltage Vp, because it is applied to a parallel combination of resistors The voltage

applied to both R2and R3is reduced by the amount V1, and so it is

Vp = V − V1= 12.0 V − 2.35 V = 9.65 V

Now the current I2through resistance R2is found using Ohm’s law:

I2= V Rp

2 = 6.00 Ω9.65 V = 1.61 A

Discussion for (c)

The current is less than the 2.00 A that flowed through R2 when it was connected in parallel to the battery in the previous parallel circuit example